Does this series converge or diverge?
Solution
I found that:
Therefore the series is divergent by comparison with harmonic series?
Thank you.
Hello,
It is supposed to converge.
It's a Bertrand series : Bertrand Series
Also note that
Make
Alright. That makes sense. How do you avoid the pitfall of what I did?
How do you come up with f(n)? Take the difference between the series you're comparing?
Thanks Mathstud for all of your posts. I'm amazed you have such a clear understanding of the material at a young age. I hope to gain this at some point, although it doesn't help when you use a crappy calculus text.
It is usually easier if the series are algebraic, for example since then
Where is some arbitrary constanst greater than one.
In this case it takes a little more intuitiveness, you have to base the fact that is a lower order infinity than any function of the form , or you can write that . A good way of thinking about it is that increases quicker than (take the derivatives and convince yourself), so eventually even if may seem to be bigger, because of 's superior derivative it will eventually "catch up" to and eventually pass it. Remember number of 's gives it quite a lot of time to do so.
Yes. If
How do you come up with f(n)? Take the difference between the series you're comparing?
Then that means that where
Thank youThanks Mathstud for all of your posts. I'm amazed you have such a clear understanding of the material at a young age. I hope to gain this at some point, although it doesn't help when you use a crappy calculus text.
Here is how I would formally prove this series is convergent.
Consider the function , differentiation yields . Solving this gives when . Now which implies that is a minimum of . Now since .
So now from that we can see that
So it follows that
And since
So by the integral test and share convergence
Thus converges
So you can model it off of that if you ever need to use it.
NOTE: A lot of people criticize me for being too verbose, you might want to cut out some uneeded words.
NOTE ALSO: This could have more easily been done by the integral test